笔记代码（各种算法实现）

//此法可以按照不排序输出，符合题意用vector<pair<string,int>>int main (){int invalid = 0;vector<pair<string,int>> VoteResult;vector<pair<string,int>>::iterator it;string str1 , str2 ;int i=0;int found=0;while(1){cin >> str1;if ( str1 == "addCandidate" ){cin >> str2 ;for (it=VoteResult.begin();it!=VoteResult.end();it++){if (it->first==str2){found=1;break;}}if(found!=1)VoteResult.push_back(make_pair(str2,0));}else if( str1 == "vote"){cin >> str2 ;found=0;for (it=VoteResult.begin();it!=VoteResult.end();it++){if (it->first==str2){found=1;break;}}if (found==1)it->second+=1;elseinvalid += 1;}else if( str1 == "getVoteResult" ){for ( it = VoteResult.begin() ; it != VoteResult.end(); it++ )cout << (*it).first << " " << (*it).second << endl;cout << invalid <<endl;return 0;}}return 0;}//map实现，升序输出候选人，不符合题意int main (){    int invalid = 0;    map<string,int> VoteResult;    map<string,int>::iterator it;    string str1 , str2 ;     while(1){cin >> str1;if ( str1 == "addCandidate" ){cin >> str2 ;VoteResult[str2] = 0;}else if( str1 == "vote"){cin >> str2 ;if ( VoteResult.find( str2 ) != VoteResult.end() )VoteResult[ str2 ] += 1; elseinvalid += 1;}else if( str1 == "getVoteResult" ){for ( it = VoteResult.begin() ; it != VoteResult.end(); it++ )cout << (*it).first << " " << (*it).second << endl;cout << invalid <<endl;return 0;}}return 0;}//一串字符，连续出现次数限制为x=3，如aaaa-bcdbint main(){string s_in;cin>>s_in;int len=s_in.size();int f=0;char s_out[100];//[1000]={'\0'};char tmp='\0';for (int i=0;i<len;i++){if (s_in[i]!=tmp){s_out[i]=(s_in[i]-'a'+1)%26+'a';tmp=s_in[i];f=0;}else{f++;if(f==1)s_out[i]=(s_in[i]-'a'+2)%26+'a';else if (f==2)s_out[i]=(s_in[i]-'a'+3)%26+'a';else{s_out[i]=(s_in[i]-'a'+1)%26+'a';f=0;}}}s_out[len]='\0';for (int i=0;i<len;i++){cout<<s_out[i];}cout<<endl;return 0;}//输出整型数据中存在指定整数的数据，并按照升序排列返回int compare (const void * a, const void * b){  return ( *(int*)a - *(int*)b );}int *calcTimes(int num[], int len,int value){vector<int> data;qsort(num,len,sizeof(int),compare);   //使用快速排序例程进行排序                                     //sort(vec.begin(),vec,end())默认升序排序，若                                         //要降序可用                                //bool cmp(const int &a,const int &b){return a>b;}  sort(vec.begin(),vec.end(),cmp);               //也可用sort升序排序完成后用，元素翻转函数reverse(vec.begin(),vec.end());int n=1,tmp=value;while(tmp!=0)  //求出tmp是几位数{n*=10;tmp/=10;}int reminder=0;for(int i=0;i<len;i++){if(num[i] < n && num[i] != value)continue;else if(num[i] == value){data.push_back(num[i]);}else{tmp = num[i];while(tmp >= value){reminder = tmp%n;if(reminder == value){data.push_back(num[i]);break;}else{tmp/=10;}}}}int *result = new int[data.size()];for( int i=0;i<data.size();i++)//这里完全是为了返回需要。。。主要是端口限制带来的麻烦{result[i] = data[i];}cout<<endl;vector<int>::iterator it;   //这是容器输出的一种方式，也可以用data[i]的方式输出for (it=data.begin();it!=data.end();it++){cout<<*it<<endl;}return result; //只返回数组的首个元素或者首元素地址}int main(){int len,value;cout<<"enter len and value";cin>>len>>value;int *data = new int[len];cout<<"enter len numbers of data[len]";for(int i=0;i<len;i++){cin>>data[i];}int *result = calcTimes(data,len,value);delete[] data;return 0;}//输入数组与指定数字，并以差值为标准升序排序输出数组，可以解决差值相等情况/*struct num{int rs;int data;}s[100];int cmp(const void *a ,const void *b){struct num* c=(num*)a;struct num* d=(num*)b;if(c->rs!=d->rs)return c->rs - d->rs;elsereturn c->data - d->data;/ *  //写法一if((*(num*)a).rs!=(*(num*)b).rs)return (*(num*)a).rs - (*(num*)b).rs;elsereturn (*(num*)a).data - (*(num*)b).data;* /}int main(){int n;int val;cout<<"shu ru n:";cin>>n;cout<<"shu ru val:";cin>>val;for (int i=0;i<n;i++){cin>>s[i].data;s[i].rs=abs(s[i].data-val);}qsort(s,n,sizeof(s[0]),cmp);for (int i=0;i<n;i++){cout<<s[i].data<<" ";}cout<<endl;return 0;}*///取出整型数据中出现次数最多的元素并升序输出/*vector<int> calTimes(int a[],int len){map<int,int>cont;map<int,int>::iterator it;vector<int>rel;vector<int>::iterator itt;int Max=INT_MIN;for (int i=0;i<len;i++){it=cont.find(a[i]);if(it==cont.end())cont[a[i]]=1;else{int t=it->second;cont.erase(it);cont[a[i]]=t+1;}}for (it=cont.begin();it!=cont.end();it++){if(it->second>Max)Max=it->second;}for (it=cont.begin();it!=cont.end();it++){if(it->second==Max)rel.push_back(it->first);}sort(rel.begin(),rel.end());return rel;}int main(){int a[10]={1,1,3,4,4,4,9,9,9,10};int len=10;vector<int>RE=calTimes(a,10);vector<int>::iterator it;for (it=RE.begin();it!=RE.end();it++){cout<<*it<<" ";}cout<<endl;return 0;}*///输入字符统计每个字符输入个数，并升序输出字符和统计个数/*int main(){map<char,int>cnt;map<char,int>::iterator it;char c[1000]={'\0'};cin>>c;int len=strlen(c);for (int i=0;i<len;i++){it=cnt.find(c[i]);if (it==cnt.end())cnt[c[i]]=1;else{int t=it->second;//保证it不是空的cnt.erase(it);cnt[c[i]]=t+1;}}for (it=cnt.begin();it!=cnt.end();it++){cout<<it->first<<" "<<it->second<<" ";}cout<<endl;return 0;}*//*******m个字符以n个一行，不足补0*****\int main(){string s;int n;cout<<"enter m zi fu:";cin>>s;cout<<"n mei gei";cin>>n;int len=s.size();for (int i=1;i<=len;i++){if(i%n!=0) cout<<s[i-1];else cout<<s[i-1]<<endl;}int yu=len%n;int number=n-yu;for(int j=1;j<=number;j++)cout<<0;cout<<endl;return 0;}\***********//*******电话号码翻译*******\int main(){char a[11][11]={"zero","one","two","three","four","five","six","seven","eight","nine","double"};char c='0';char tmp[11];int f,d;while (c!='\n'){scanf("%s%c",&tmp,&c);f=0;for (int i=0;i<11;i++){if (!strcmp(tmp,a[i])&&i<10){f=1;cout<<i;if (d==1){cout<<i;d=0;}}else if (!strcmp(tmp,a[i])&&i==10){d=1;f=1;}}if(f==0) break;}if(f==0||d==1||f==2) cout<<"Error"<<endl;return 0;}/**********频率数组记录出现次数最多的字符*******\int main(){string s;cout<<"shu ru zi fu chuan:";cin>>s;int len=s.size();int cnt[200];int max_cnt=-1;memset(cnt,0,sizeof(cnt));for (int i=0;i<len;i++){int num=s[i];//将字符转换为ascii码值cnt[num]++;if (cnt[num]>max_cnt)max_cnt=cnt[num];//max_cnt=max(max_cnt,cnt[s[i]]);}for (int j=0;j<=129;j++){if (cnt[j]==max_cnt){cout<<(char)j<<endl;break;//为了保证输出的是第一个最小的字典序字符}}return 0;}\**************//**********以整型数输入方式*****\  int cal(int n)//计算各个位上的和//输出按各位和升序，若和相等则按数值大小升序排序{int ans=0;if(n==0)return 0;else{while (n!=0){ans+=n%10;n/=10;}return ans;}}int cmp(int x,int y){if(cal(x)!=cal(y))return cal(x)<cal(y);elsereturn x<y;}int main(){int arr[10];for (int i=0;i<10;i++){cin>>arr[i];}sort(arr,arr+10,cmp);for (int i=0;i<10;i++){if(i==9)cout<<arr[i];elsecout<<arr[i]<<" ";}return 0;}\************///**//\\\**\给定的字符串中找单词，按长度降序排列，并输入到新字符串中，空格，重复只输出一次//容器法#include<iostream>#include<string>#include<vector>#include<algorithm>using namespace std;bool myfunction (const string &str1,const string &str2) { return (str2.length() < str1.length());}void my_word(char input[], char output[]){vector<string> str;char *cur = input;char *p = input;int len;bool flag = false;//去除杂质，取单词while(*p != '\0'){if((*p >= 'a' && *p <= 'z') || (*p >= 'A' && *p <= 'Z'))p++;else{len = p-cur;if(len > 1){string tmp(cur,len);for(int i=0;i<str.size();i++)if(tmp == str[i])flag = true;if(flag != true)str.push_back(tmp);}cur = p+1;p++;flag = false;}}len = p-cur;if(len > 1){string tmp(cur,len);for(int i=0;i<str.size();i++)if(tmp == str[i])flag = true;if(flag != true)str.push_back(tmp);}//sortsort(str.begin(),str.end(),myfunction);//拼接字符串*output='\0';int i;for(i=0;i<str.size()-1;i++){strcpy(output+strlen(output),(str[i]+" ").c_str());}strcpy(output+strlen(output),(str[i]).c_str());cout<<output<<endl;}int main(){char input[]="some ninja local buses, a bd b c d somes1234123drivers";char output[500];my_word(input, output);return 0;}///注意char**指针的传递通过键盘输入一串小写字母(a~z)组成的字符串。请编写一个字符串过滤程序，若字符串中出现多个相同的字符，将非首次出现的字符过滤掉。比如字符串"abacacde"过滤结果为"abcde"。要求实现函数： void stringFilter(const char *pInputStr, long lInputLen, char *pOutputStr);【输入】 pInputStr：  输入字符串         lInputLen：  输入字符串长度         【输出】 pOutputStr： 输出字符串，空间已经开辟好，与输入字符串等长；【注意】只需要完成该函数功能算法，中间不需要有任何IO的输入输出示例 输入："deefd"        输出："def"输入："afafafaf"     输出："af"输入："pppppppp"     输出："p"//20:29 2013/7/24// main函数已经隐藏，这里保留给用户的测试入口，在这里测试你的实现函数，可以调用printf打印输出// 当前你可以使用其他方法测试，只要保证最终程序能正确执行即可 // 该函数实现可以任意修改，但是不要改变函数原型。一定要保证编译运行不受影响void stringFilter(const char *pInputStr, long lInputLen, char *pOutputStr){}*/#include "iostream"#include "cstring"#include "malloc.h"using namespace std;char* stringFilter(const char *pInputStr, long lInputLen, char *pOutputStr){char *cur = NULL;char *r = pOutputStr;*pOutputStr='\0';while(*pInputStr != '\0'){cur = pOutputStr; while(*cur != '\0'){if(*cur == *pInputStr)break;cur++;}if(*cur == '\0')*(r++) = *(pInputStr++);elsepInputStr++;}*r='\0';return pOutputStr;}int main(){//const char*a[]=="aabfcadfakeafccdede";char*a="aabfcadfakeafccdede";cout<<"过滤前字符串："<<a<<endl;long len=strlen(a);char *b=(char*)malloc(len*sizeof(char));b=stringFilter(a,len,b);cout<<"过滤后字符串：";cout<<b<<endl;return 0;}//filter程序int main(){string s;cin>>s;int len=s.size();int a[200]={0};char b[200]={'0'};int j=0;for (int i=0;i<len;i++){if (a[s[i]-'0']==0){b[j]=s[i];j++;a[s[i]-'0']=1;}elsecontinue;}for (int i=0;i<j;i++){cout<<b[i];}cout<<endl;return 0;}/*通过键盘输入一串小写字母(a~z)组成的字符串。请编写一个字符串压缩程序，将字符串中连续出席的重复字母进行压缩，并输出压缩后的字符串。压缩规则：1. 仅压缩连续重复出现的字符。比如字符串"abcbc"由于无连续重复字符，压缩后的字符串还是"abcbc".2. 压缩字段的格式为"字符重复的次数+字符"。例如：字符串"xxxyyyyyyz"压缩后就成为"3x6yz"要求实现函数： void stringZip(const char *pInputStr, long lInputLen, char *pOutputStr);【输入】 pInputStr： 输入字符串lInputLen： 输入字符串长度 【输出】 pOutputStr： 输出字符串，空间已经开辟好，与输入字符串等长；*/#include <stdio.h>#include<string.h>#include<malloc.h>void stringZip(const char *p_str, long len, char *p_outstr){int count=1;for(int i=0;i<len;i++){if(p_str[i]==p_str[i+1]){count++;}else{if(count>1){*p_outstr++ = count +'0';*p_outstr++ =p_str[i];}else{*p_outstr++ =p_str[i];}count = 1;//注意其位置}}*p_outstr = '\0';}void main(){char *str = "cccddeccffgghhhhjhkk";    printf("压缩之前的字符串为：%s\n",str);int len = strlen(str);char * outstr = (char*)malloc(len*sizeof(char));stringZip(str,len,outstr);printf("压缩之后的字符串为：%s\n",outstr);free(outstr);outstr = NULL;}//map<char ,int>实现字符压缩程序/*int main(){map<char,int> zip;map<char,int>::iterator it;string s;cin>>s;int len=s.size();for (int i=0;i<len;i++){it=zip.find(s[i]);if (it==zip.end()){zip[s[i]]=1;}else{int t= it->second;zip.erase(it);zip[s[i]]=t+1;}}for (it=zip.begin();it!=zip.end();it++){cout<<it->second<<it->first<<"  ";}cout<<endl;return 0;}*////////////////////////////////////////////////////////删除要查找的字符串，并返回删除次数int delete_sub_str(char *input,char *sub_str,char *result){int count=0;int len = strlen(sub_str);char *p = result;while(*input != '\0'){int i=0;for(;i<len;i++){*p = *input;if(*input != *(sub_str+i)){p++;input++;break;;}p++;input++;}if(i == len){p = p-len;count++;}}*p='\0';return count;}////////////////////////////////////////////////////////////////字串分离，并以逗号隔开#include "iostream"#include "cstring"#include "malloc.h"using namespace std;void DivideString(const char* pInputStr , long lInputLen , char* pOutputStr)  {    int i , j;    bool flag;      for(i = 0 ; pInputStr[i] == ' ' ; ++i)   //跳过字符串前面的空格          ;      flag = true;      for(j = 0 ; i < lInputLen ; ++i)      {          if(pInputStr[i] != ' ')         {              if(!flag)                 flag = true;              pOutputStr[j++] = pInputStr[i];    //将各个子串分离保存下来          }          else         {              if(flag)                  pOutputStr[j++] = ',';              flag = false;          }      }      pOutputStr[j++] = ',';      pOutputStr[j++] = '\0';  }  int main(){char *input="abcd efg hijkl mn opq      china";int len=strlen(input);char* output=(char*)malloc(len*sizeof(char));DivideString(input,len,output);cout<<output<<endl;return 0;}//filter 的牛逼写法void stringFilter(const char *pInputStr, long lInputLen, char *pOutputStr){int a[26] = {0};int pos = 0;long i, j;const char *p = pInputStr;for(i = 0, j = 0; i < lInputLen; i++){pos = pInputStr[i] - 'a'; //将出现的字符标记为1，下次出现同样的字符时就不再存储if(a[pos] == 0){a[pos]++;pOutputStr[j++] = pInputStr[i];}}pOutputStr[j] = '\0';}//压缩程序的牛逼写法void stringZip(const char *pInputStr, long lInputLen, char *pOutputStr){int i, j, k = 0, cnt = 0;char buffer[30];for(i = 0; i < lInputLen;){for(j = i + 1;;j++){if(pInputStr[j] == pInputStr[i])cnt++;elsebreak;}if(cnt != 0){memset(buffer, 0, sizeof(buffer));itoa(cnt + 1, buffer, 10);strcat(pOutputStr, buffer);k += strlen(buffer);i += cnt;}pOutputStr[k++] = pInputStr[i++];cnt = 0;}}//阿里巴巴第一题字符串查找#include <stdio.h>#include <string.h>int FindQueryInText( char query[], char text[]);     //查找长度函数void GetQuery( char sourcequery [], char densquery[],int startIndex,int querylength );//从指定位置读取指定长度的到目的字符串中void main()          //测试main函数{int length = 0;               //运算结果长度char query[] = "acbac";       //初始化字符串char text[] = "acaccbabb";    //初始化被查找字符串length = FindQueryInText( query,text);  //调用函数给出结果printf("%d",length);}int FindQueryInText( char query[], char text[]){int querylength = 0,textlength = 0;int i = 0,j=0;char comparequery[100] = {0};   char *pqueryPosition = NULL;querylength = strlen(query);textlength = strlen(text);for(i = querylength;i>0;i--)          //循环querylength次，依次查找是否有字符串{for(j = 0;j<querylength-i+1;j++)  //每次查找要从query中取querylength-i次{GetQuery(query,comparequery,j,i); //每次比较从query中取出i个字符，进行比较pqueryPosition=strstr(text,comparequery);//比较，若是text中用comparequery中，则返回非NULL指针。if (pqueryPosition != NULL){break;}}if (pqueryPosition != NULL){break;}}return i;}void GetQuery( char sourcequery [], char densquery[],int startIndex,int querylength ){int i = 0;for (i = 0;i < querylength;i ++ ){densquery[i] = sourcequery[i+startIndex];}densquery[querylength] = '\0';}//写法二阿里巴巴第一题#include <stdio.h>  #include <string.h>   using namespace std;#define qurey "acbac"#define text  "acaccbabb"#define N 10002 int c[N][N];int STRICT_LCS_LENGTH(char *X, char *Y,int m,int n){int i,j,max=0;for ( i = 0;i < m;i++)c[i][0] = 0;for ( j = 0;j < n;j++){c[0][j] = 0;}for( i = 0;i < m;i++){for ( j = 0;j < n;j++){if (X[i] == Y[j]){c[i][j] = c[i-1][j-1]+1;if(c[i][j] > max)max = c[i][j];}else c[i][j] = 0;}}return max;}int main(){  char a[N] = qurey;char b[N] = text;int m = strlen(a);int n = strlen(b);int len = STRICT_LCS_LENGTH(a,b,m,n);printf("%d\n",len);}  //西安交大第一题，第二字符串补齐到第一字符串（定长）并输出/*给定三个参数，src，len，str，其中src，和str为字符窜，将str中内容补齐到src后，补齐后总长为len如输入：src="abc",str="12345" len=8   输出：abc12345;如输入：src="abc",str="1234567" len=8   输出：abc12345;如输入：src="abc",str="12" len=8   输出：abc12121;*/#include "iostream"#include "cstring"using namespace std;int main(){char src[100],str[100];cin>>src>>str;int src_len=strlen(src);int str_len=strlen(str);/*  方法一if (str_len>=8-src_len){str[8-src_len]='\0';strcat(src,str);}else if (str_len<8-src_len){int j=8-src_len;for (int i=0;i<j;i++){int n=i;n=n%str_len;src[src_len+i]=str[n];}src[8]='\0';}*///方法二int j=8-src_len;for (int i=0;i<j;i++){int n=i;n=n%str_len;src[src_len+i]=str[n];}src[8]='\0';cout<<src<<endl;return 0;}//西安交大第二题国名排序#include "iostream"#include "cstring"#include "string"using namespace std;int main(){string s[5];for (int i=0;i<5;i++){cin>>s[i];}for (int i=0;i<5;i++){for (int j=i+1;j<5;j++){if (s[i]>s[j]){swap(s[i],s[j]);}}}for (int k=0;k<5;k++){cout<<s[k]<<" ";}cout<<endl;return 0;}/*******函数实现选择法排序数组*************\int main(){void select_sort(int array[],int n);int a[10]={12,43,34,6,73,123,123,45,2354,2};select_sort(a,10);for (int i=0;i<10;i++){cout<<a[i]<<endl;}return 0;}void select_sort(int array[],int n){int i,j,k,t;for (i=0;i<n-1;i++){k=i;for (j=i+1;j<n;j++){if(array[j]<array[k])k=j;t=array[k];array[k]=array[i];array[i]=t;}//return;}}将整数倒序输出，剔除重复数据输入一个整数，如12336544，或1750，然后从最后一位开始倒过来输出，最后如果是0，则不输出，输出的数字是不带重复数字的，所以上面的输出是456321和571。如果是负数，比如输入-175，输出-571。*/#include "iostream"using namespace std;int main(){int Num,n[50],a[10]={0},temp,i=0,flag=0;char c;scanf("%c%d",&c,&Num);while (Num>0){temp=Num%10;if (a[temp]==0){a[temp]=1;n[i]=temp;i++;}Num=Num/10;}if(c=='-') printf("%c",c);for (temp=0;temp<i;temp++){if ((n[temp]!=0)||flag!=0){printf("%d",n[temp]);flag=1;}}if(c!='-') printf("%d",(int)c-48);return 0;}整数化为2进制数，并逆序输出*/#include "iostream"using namespace std;int main(){int m;cout<<"输入要转化的正整数：";cin>>m;int a[32]={0},i=0,j;while (m>0){a[i]=m%2; //如果转换成16进制，可以将m改为         //M[m%16]的表，M表中存0~F字符         //并且a[]也要定义成char型数据i++;m/=2;}for (j=0;j<i;j++){cout<<a[j];}cout<<"正序为：";for (j=i-1;j>=0;j--){cout<<a[j];}return 0;}//一串字符串分成多字子串，以空格隔开#include "iostream"#include "string.h"#include "stdlib.h"using namespace std;int main(){char a[]="abcd,efg Hijk/lmnop0qrst Uvwxyz";char b[]="def";int len=strlen(a);for (int i=0;i<len;i++)  {  if (a[i]<'A' || (a[i]>'Z'&&a[i]<'a') || a[i]>'z')  {  a[i]=',';  }  }  char *p;char except[]=",";int j=0;char *word[100];p=strtok(a,except);while (p!=NULL){word[100]=p;cout<<p<<endl;p=strtok(NULL,except);cout<<"word"<<j<<"wei:"<<word<<endl;} 大数相减输入两行字符串正整数，第一行是被减数，第二行是减数，输出第一行减去第二行的结果。备注：1、两个整数都是正整数，被减数大于减数示例：输入：1000000000000001      1输出：1000000000000000注意大数用char a[] 存储，用%s接收，一位一位的运算。注意a[0]里的正负号。*/#include "iostream"using namespace std;int main(){char s1[1010]={'\0'},s2[1010]={'\0'};scanf("%s%s",&s1,&s2);int len1=strlen(s1),len2=strlen(s2),i,j,m1,m2,f=0;int answer[1010],k=0;for (i=len1-1,j=len2-1;i>=0;i--,j--){m1=(int)s1[i]-48;if(j>=0) m2=(int)s2[j]-48;else m2=0;if(f==1) //先判断是否有借位，然后在计算m1-m2{m1=m1-1;f=0;}if(m1>=m2) {answer[k++]=m1-m2;f=0;}else {answer[k++]=m1+10-m2;f=1;}}int t=0;//帮助判断高位是否为0for (k=k-1;k>=0;k--){if(t==0&&answer[k]==0) continue;else {t=1;printf("%d",answer[k]);}}return 0;}将 电话号码 one two 。。。nine zero翻译成1  2 。。9 0中间会有double例如输入：OneTwoThree输出：123输入：OneTwoDoubleTwo输出：1222输入：1Two2 输出：ERROR输入：DoubleDoubleTwo 输出：ERROR第三题：有空格，非法字符，两个Double相连，Double位于最后一个单词 都错误*/#include "iostream"using namespace std;int main(){char a[11][11]={"zero","one","two","three","four","five",            "six","seven","eight","nine","double"};char temp[11];char c=' ';int i,d=0,f;cout<<"输入数字格式如（one two double three）:";while (c!='\n'){scanf("%s%c",&temp,&c);f=0;for (i=0;i<11;i++){if (!strcmp(temp,a[i])&&i<10){printf("%d",i);f=1;if (d==1){printf("%d",i);d=0;}}else if (!strcmp(temp,a[i])&&i==10){d=1;f=1;}}if(f==0) break;}printf("\n");if (d==1||f==0)printf("Error\n");return 0;}字符串M化成以N为单位的段输入m个字符串 和一个整数n, 把字符串M化成以N为单位的段，不足的位数用0补齐。如 n=8 m=9 ，123456789划分为：1234567890000000123化为 ：12300000*/#include "iostream"using namespace std;int main(){int m,n;char str[100]={'\0'};int len;int yu,shang;printf("输入字符窜字符个数：");scanf("%d",&m);scanf("%s",&str);len=strlen(str);printf("输入n的值：");scanf("%d",&n);yu=len%n;for (int i=0;i<len;i++){if(i%n==0)printf("\n%c",str[i]);else printf("%c",str[i]);}if (yu==0);else{for(int i=yu+1;i<=n;i++)printf("0");printf("\n");}return 0;}//子串合并/*int main(){char *c="   abc def gh i      d";int len =strlen(c);char *out=(char*)malloc(len*sizeof(char));bool flag;int j;int i;for (i=0;c[i]==' ';i++);for (j=0;i<len;i++){if (c[i]!=' '){flag=true;out[j++]=c[i];}else{if(flag)out[j++]=',';flag=false;}}out[j++]=',';out[j++]='\0';for (int i=0;i<j;i++){cout<<out[i];}cout<<endl;return 0;}*///strstr的实现写法#include "iostream"using namespace std;char* strstra(const char* src,const char* pat){char* temp1= (char*)src;char* temp2= (char*)pat;char *temp3;if(!*temp2) return temp1;while(*temp1!='\0'){temp3 =  temp1;while(*temp3 && *temp2 && !(*temp1-*temp2)){temp3++;temp2++;}if(!*temp2) return temp1;temp1++;}return NULL;}int main(){char *s1="abcdefj";char *s2="def";char*c=strstra(s1,s2);cout<<c<<endl;return 0;}//10进制转换成x进制/***************\int x=8;int i=0;char num[100]={0};int number;cin>>number;while (number!=0){if ((number%x)>=10&&(number%x<=15)){num[i]=(char)((number%x)+55);//10-x转换i++;number/=x;}else if (number%x<10){num[i]=(number%x)+'0';//10-x转换i++;number/=x;}}for (int j=i-1;j>=0;j--)//按阅读习惯输出{cout<<num[j];}\**************************//****\//x-10转换string s;cin>>s;double num=0;double x=2;int len=s.size();         char c=9;//可以直接将 9 赋值给char，此时cout<<c 无输出显示空白         cout<<c<<endl;         int a=(c-'0')*2;//字符可以直接参与运算并赋值给整型数据的         cout<<a<<endl;double j=0;for (int i=len-1;i>=0;i--)  //字符串输入，借助pow函数完成x-10的转换，注意pow的数据和返回值都是double型{if (s[i]>='A'&&s[i]<='F'){num=num+((s[i]-'A')+10)*pow(x,j);  j++;}else if (s[i]>='a'&&s[i]<='f'){num=num+((s[i]-'a')+10)*pow(x,j);  //可以用数学函数pow，但要求是double型数据j++;}else if (s[i]>='0'&&s[i]<='9'){num=num+(s[i]-'0')*pow(x,j);  //可以用数学函数pow，但要求是double型数据j++;}}cout<<num;/////菲薄拉塞序列输出cout<<"******************************************加强版Fibonacci数序写法"<<endl;//int main(){int i;int f[20]={1,1};for (i=2;i<20;i++){f[i]=f[i-1]+f[i-2];}for (i=0;i<20;i++){if(i%4==0) cout<<endl;cout<<setw(8)<<f[i];}cout<<endl;}/////////////////////////////列车站台问题/*#include<stdio.h>#include<string>#include<queue>#include<vector>using namespace std;#define MAX 35#define SUBWAY_A 20#define SUBWAY_B 15typedef struct node{int adjvex;struct node *next;}edgenode;typedef struct{char name[10];bool flag;edgenode *link;}vexnode;const char subway_name1[SUBWAY_A][10]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1","A10","A11","A12","A13","T2","A14","A15","A16","A17","A18"};const char subway_name2[SUBWAY_B][10]={"B1","B2","B3","B4","B5","B6","B7","B8","B9","B10","B11","B12","B13","B14","B15"};void creat(vexnode ga[]){int i;edgenode *p;for(i=0;i<MAX;i++){ga[i].link=NULL;ga[i].flag=true;if(i<SUBWAY_A) strcpy(ga[i].name,subway_name1[i]);else strcpy(ga[i].name,subway_name2[i-20]);}//A地铁建邻接表for(i=1;i<SUBWAY_A-1;i++){p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=i-1;p->next=NULL;ga[i].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=i+1;p->next=NULL;ga[i].link->next=p;if(i==9){p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+4;p->next=NULL;ga[i].link->next->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+5;p->next=NULL;ga[i].link->next->next->next=p;}else if(i==14){p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+9;p->next=NULL;ga[i].link->next->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+10;p->next=NULL;ga[i].link->next->next->next=p;}}p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A-1;p->next=NULL;ga[0].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=1;p->next=NULL;ga[0].link->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A-2;p->next=NULL;ga[SUBWAY_A-1].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=0;p->next=NULL;ga[SUBWAY_A-1].link->next=p;//B地铁建邻接表for(i=1;i<SUBWAY_B-1;i++){if(i==4||i==5||i==9||i==10) continue;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+i-1;p->next=NULL;ga[i+SUBWAY_A].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+i+1;p->next=NULL;ga[i+SUBWAY_A].link->next=p;}p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+3;p->next=NULL;ga[SUBWAY_A+4].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=9;p->next=NULL;ga[SUBWAY_A+4].link->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=9;p->next=NULL;ga[SUBWAY_A+5].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+6;p->next=NULL;ga[SUBWAY_A+5].link->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+8;p->next=NULL;ga[SUBWAY_A+9].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=14;p->next=NULL;ga[SUBWAY_A+9].link->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=14;p->next=NULL;ga[SUBWAY_A+10].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+11;p->next=NULL;ga[SUBWAY_A+10].link->next=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+1;p->next=NULL;ga[SUBWAY_A].link=p;p=(edgenode*)malloc(sizeof(edgenode));p->adjvex=SUBWAY_A+SUBWAY_B-2;p->next=NULL;ga[SUBWAY_A+SUBWAY_B-1].link=p;//  打印各邻接节点for(i=0;i<MAX;i++){printf("%s:",ga[i].name);edgenode *s;s=ga[i].link;while(s!=NULL){printf("->%s",ga[s->adjvex].name);s=s->next;}printf("\n");}}int main(){vexnode ga[MAX];creat(ga);int i;char str[2][10];while(scanf("%s%s",str[0],str[1])!=EOF){int temp=0;for(i=0;i<MAX;i++){ga[i].flag=true;if(!strcmp(str[0],ga[i].name)) temp=i;}queue<vexnode>q;q.push(ga[temp]);ga[temp].flag=false;int count=0;int start=0;int end=1;bool find_flag=false;while(!q.empty()){if(find_flag) break;count++;printf("************************\n");printf("第%d层搜索：",count);int temp_end=end;                while(start<temp_end){printf("%s ",q.front().name);if(!strcmp(q.front().name,str[1])){find_flag=true;break;}edgenode *s;s=q.front().link;while(s!=NULL){if(ga[s->adjvex].flag){q.push(ga[s->adjvex]);ga[s->adjvex].flag=false;end++;//printf("%s ",ga[s->adjvex].name);}s=s->next;                }q.pop();start++;}printf("\n");}printf("%d\n",count);}return 0;}*///写法二站台问题#include <cstdlib>  #include <iostream>  #include <string>/* Author : 俗人Time ： 2013/9/2 description : 地铁换乘问题 已知2条地铁线路,其中A为环线,B为东西向线路,线路均为双向,换乘点为 T1,T2,编写程序,任意输入两个站点名称,输出乘坐地铁最少所需要经过的车站数量A(环线):A1...A9 T1 A10...A13 T2 A14...A18 B(直线):B1...B5 T1 B6...B10 T2 B11...B15 样例输入:A1 A3样例输出:3 */   using namespace std;//无向图的数据结构  struct Graph  {         int arrArc[100][100];  //邻接矩阵        int verCount;  //点数        int arcCount;  //边数        };  //FLOYD算法 求任意两点最短路径矩阵       void floyd(Graph *p,int dis[100][100]){              for(int k = 1;k <= p->verCount;k++)          for(int i = 1; i <= p->verCount;i++)             for(int j = 1;j <= p->verCount;j++)             {                     //存在更近的路径，则更新                       if(dis[i][j]>dis[i][k]+dis[k][j])                     dis[i][j]=dis[i][k]+dis[k][j];             }              }            //站名字符串转节点编号     int  char_to_int(string s){              string s1[38] = {"A0","A1","A2","A3","A4","A5","A6","A7","A8","A9","A10",                    "A11","A12","A13","A14","A15","A16","A17","A18",                    "B1","B2","B3","B4","B5","B6","B7","B8","B9","B10",                    "B11","B12","B13","B14","B15","T1","T2"} ;         for(int i=1 ; i <= 38;i ++){                        if (s == s1[i])            return i;                 }          return -1;            }      int main(int argc, char *argv[])  {      Graph g;          g.verCount = 37;      g.arcCount = 36;        cout<<"number of ver:"<<g.verCount<<" number of arc:" <<g.arcCount<<endl;          //初始化邻接矩阵        for(int i = 1;i<=g.verCount;i++)      { for(int j = 1;j<=g.verCount;j++)         {                //i到本身的距离为0                 //不同节点值为不可达                  if(i==j) g.arrArc[i][i]= 0;                else                g.arrArc[i][j] = 65535;         }      }                   //输入A环线个点相连情况 每个边权重都为1             int a[21] = {1,2,3,4,5,6,7,8,9,34,10,11,12,13,35,14,15,16,17,18,1};        for(int i=0;i<20;i++)      {          g.arrArc[a[i]][a[i+1]] = 1;         g.arrArc[a[i+1]][a[i]] = 1;          }          //输入B线个点相连情况 每个边权重都为1     int b[17] = {19,20,21,22,23,34,24,25,26,27,28,35,29,30,31,32,33};       for(int i=0;i<16;i++)      {           g.arrArc[b[i]][b[i+1]] = 1;         g.arrArc[b[i+1]][b[i]] = 1;            }         //计算邻接矩阵     floyd(&g,g.arrArc);              cout << "请输入起始站点与终点:" <<endl;     string start;     string end;           cin >> start >> end;    cout << g.arrArc[char_to_int(start)][char_to_int(end)]+1 <<endl;        system("PAUSE");        return EXIT_SUCCESS;  }//两个正整数相减或相加并输出结果#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAXCHAR 256//必须规定均为小写字母void stringFilter(const char *pInputStr, long lInputLen, char *pOutputStr){int a[26] = {0};int pos = 0;long i, j;const char *p = pInputStr;for(i = 0, j = 0; i < lInputLen; i++){pos = pInputStr[i] - 'a'; //将出现的字符标记为1，下次出现同样的字符时就不再存储if(a[pos] == 0){a[pos]++;pOutputStr[j++] = pInputStr[i];}}pOutputStr[j] = '\0';}void stringZip(const char *pInputStr, long lInputLen, char *pOutputStr){int i, j, k = 0, cnt = 0;char buffer[30];for(i = 0; i < lInputLen;){for(j = i + 1;;j++){if(pInputStr[j] == pInputStr[i])cnt++;elsebreak;}if(cnt != 0){memset(buffer, 0, sizeof(buffer));itoa(cnt + 1, buffer, 10);strcat(pOutputStr, buffer);k += strlen(buffer);i += cnt;}pOutputStr[k++] = pInputStr[i++];cnt = 0;}}void arithmetic(const char *pInputStr, long lInputLen, char *pOutputStr){int i, cnt = 0, a, b, result;char ch[1] = {'0'};char op1[MAXCHAR], op[MAXCHAR], op2[MAXCHAR], buffer[4];for(i = 0; i < lInputLen; i++)if(pInputStr[i] == ' ')cnt++;if(cnt != 2)                    //空格数不等于2{strcat(pOutputStr, ch);return;}sscanf(pInputStr, "%s %s %s", op1, op, op2);if(strlen(op) > 1 || (op[0] != '+' && op[0] != '-'))        // 操作符有多个{strcat(pOutputStr, ch);return;}for(i = 0; i < strlen(op1); i++)                            //操作数1是否有其他字符{if(op1[i] < '0' || op1[i] > '9'){    strcat(pOutputStr, ch);return;}}for(i = 0; i < strlen(op2); i++)                            //操作数2是否有其他字符{if(op2[i] < '0' || op2[i] > '9'){strcat(pOutputStr, ch);return;}}a = atoi(op1);b = atoi(op2);switch(op[0]){case '+':result = a + b;itoa(result, buffer, 10);strcat(pOutputStr, buffer);break;case '-':result = a - b;itoa(result, buffer, 10);strcat(pOutputStr, buffer);break;default:break;}}int main(){char pInputStr1[] = {"aaabbbcccdde"};char pInputStr2[] = {"aaabbcddde"};char pInputStr3[] = {"3 + 4"};char pOutputStr1[MAXCHAR] = {0};char pOutputStr2[MAXCHAR] = {0};char pOutputStr3[MAXCHAR] = {0};/* TODO: 调用被测函数 */stringFilter(pInputStr1, strlen(pInputStr1), pOutputStr1);stringZip(pInputStr2, strlen(pInputStr2), pOutputStr2);arithmetic(pInputStr3, strlen(pInputStr3), pOutputStr3);/* TODO: 执行完成后可比较是否是你认为正确的值 */printf(pOutputStr1); //abcdecout<<endl;printf(pOutputStr2); //3a3b3c2decout<<endl;printf(pOutputStr3); //7cout<<endl;return 0;}